[Leet Code] Longest Continuous Increasing Subsequence

Problem:

Given an unsorted array of integers nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.

A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].

Example 1:

Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
4.

Example 2:

Input: nums = [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.

Constraints:

• 0 <= nums.length <= 104
• -109 <= nums[i] <= 109

Solution:

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        last_number = 0
        counter = 0
        max_counter = 0
        for i in range(len(nums)):
            if i == 0 or len(nums) == 0:
                counter = 1
            elif nums[i] > last_num:
                counter += 1
            else:
                counter = 1
            if counter >= max_counter:
                max_counter = counter
            last_num = nums[i]
        return max_counter